Question

form the quadratic polynomial whose zeroes are 1/a and 1/b

Answer 1

x² - (sum of zeroes)x + product of zeroes

x² - (1/a + 1/b )x + 1/a × 1/b

x² - ((b+a)/ab)x + 1/ab

${x}^{2} - \frac{b + a}{ab} x + \frac{1}{ab}$

Answer 2

$x^2 - \frac{a+b}{ab}x + \frac{1}{ab} = 0$  is the quadratic polynomial whose zeroes are 1/a and 1/b

Solution:

The general quadratic equation is:

$x^2 - ( \text{ sum of zeros } )x + \text{ product of zeros } = 0$

From given,

$Zeros = \frac{1}{a} \text{ and } \frac{1}{b}$

Find sum of zeros:

$\text{Sum of zeros } = \frac{1}{a} + \frac{1}{b}\\\\\text{Sum of zeros } = \frac{a+b}{ab}$

Find product of zeros:

$\text{Product of zeros } = \frac{1}{a} \times \frac{1}{b} = \frac{1}{ab}$

Therefore, the quadratic equation is given as:

$x^2 - \frac{a+b}{ab}x + \frac{1}{ab} = 0$

Thus the quadratic polynomial is found

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